已知三解形ABC的外接圆半径为R,且2R(sin^2A-sin^2C)=(根号2a-b)sinB,求角C.
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正弦定理:a/sinA=b/sinB=c/sinC=2Ra=2RsinA; b=2RsinB; c=2RsinC.在原等式的两边同时乘2R得到:(2R)^2[(sinA)^2-(sinC)^2]=(a√2-b)*2RsinB---(2RsinA)^2-(2RsinC)^2=(a√2-b)(2RsinB)---a^2-c^2=(a√2-b)b---a^2-c^2=ab√2-b^2.---a^2+b^2-c^2=ab√2---(a^2+b^2-c^2)/(2ab)=√2/2---cosC=√2/20---C=45°